极限求解lim[(2/π)arctanx]^x (x趋向于正的无穷大）

---

x→+∞
lim [(2/π)arctanx]^x
=lim e^ln [(2/π)arctanx]^x
=e^lim ln[(2/π)arctanx]^x
考虑
lim ln[(2/π)arctanx]^x
=lim x * ln[(2/π)arctanx]
=lim ln[1+(2/π)arctanx-1] / (1/x)
=lim [(2/π)arctanx-1] / (1/x)
该极限旦饥为0/0型，根据L'Hospital法则
=lim [(2/π)arctanx-1]' / (1/x)'
=lim (2/π)/亩斗(x^2+1) / (-1/x^2)
=(-2/π)*lim (x^2)/(1+x^2)
=-2/π
因此，原极限迅迟磨=e^(-2/π)
有不懂欢迎追问

---

解：
y=[(2/π)arctanx]^x
两边尘桐中同时取自然对数得：
lny=xln[(2/派山π)arctanx]
lim【x→+∞】lny
=lim【x→+∞】xln[(2/π)arctanx]
=lim【x→+∞】ln[(2/π)arctanx]/(1/x)
=lim【x→+∞轮蔽】1/[(2/π)arctanx]・1/(1+x²)/(-1/x²)
=-lim【x→+∞】x²/(1+x²)・1/[(2/π)arctanx]
=-1・1/[(2/π)・(π/2)]
=-1
所以lim【x→+∞】[(2/π)arctanx]^x=1/e

---